HDU-5636(Shortest Path)（floyd最短路径）

HDU-5636(Shortest Path)（floyd最短路径）

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1393    Accepted Submission(s): 440

Problem Description

There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to n and there is an edge with unit length between i and i+1(1≤i<n). To make the graph more interesting, someone adds three more edges to the graph. The length of each new edge is1.

You are given the graph and several queries about the shortest path between some pairs of vertices.

 

Input

There are multiple test cases. The first line of input contains an integerT, indicating the number of test cases. For each test case:

The first line contains two integer n and m(1≤n,m≤105) -- the number of vertices and the number of queries. The next line contains 6 integersa1,b1,a2,b2,a3,b3(1≤a1,a2,a3,b1,b2,b3≤n), separated by a space, denoting the new added three edges are (a1,b1),(a2,b2),(a3,b3).

In the next m lines, each contains two integers si and ti(1≤si,ti≤n), denoting a query.

The sum of values of m in all test cases doesn't exceed 106.

 

Output

For each test cases, output an integer S=(∑i=1mi⋅zi) mod (109+7), where zi is the answer for i-th query.

 

Sample Input

1 10 2 2 4 5 7 8 10 1 5 3 1

 

Sample Output

7

 

注：由于此题所给的点比较多，直接用最短路径肯定会超时。思考很久无果，于是看了别人的题解，刚看时，毫无头绪，硬着头皮看了很久才看懂。

特殊处理+floyd。

可以把基本情况看做：h  x[i]  x[j]   l   ，其中h与l是待求的两点，x[i]、x[j]，是6个点中的任意两个点。在求待求两点之间的最短路径时，可以把整条路径看做3段。 （1）h  到 x[i]  （2）x[i]到 x[j]   (3)  x[j] 到 l  。通过遍历所有情况来获取最短的路径。因为这六个点之间因为3座桥，而使路径变短。所以在从h到i的路径中，有桥，则可以使路径变短。

当情况为：h  l  x[i]  x[j]或x[i]  x[j]  l   h时，最短路径s，则为：s=abs(h-l);

for(i=1;i<=6;i++)
for(j=1;j<=6;j++)
{    
   ans=min(ans,abs(h-x[i])+m1[i][j]+abs(l-x[j])) ; //h与l,x[i]与x[j]的前后关系不能确定，所以需要把h与l的位置互换来确定最小值 
   ans=min(ans,abs(l-x[i])+m1[i][j]+abs(h-x[j])) ;
}

My  solution：

/*2016.3.17*/

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; 
#define mod  1000000007
long  long m1[7][7];
long long  ans;
int x[7],n,m;
void floyd()//更新6个点之间的最短距离 
{
	int i,j,k;
    for(k=1;k<=6;k++)
	for(i=1;i<=6;i++)
	for(j=1;j<=6;j++)
	{
		if(m1[i][j]>(m1[i][k]+m1[k][j]))
		m1[i][j]=m1[i][k]+m1[k][j];
	}	
	return ;
}
int main()
{
	int i,j,k,h,l,p,t;
	long long   sum,q;
	scanf("%d",&t);
	while(t--)
	{
		q=1;
		sum=0;
		scanf("%d%d",&n,&m);
		for(i=1;i<=6;i++)
		{
			scanf("%d",&x[i]);
		}
		for(i=1;i<=6;i++)
		{
			for(j=1;j<=6;j++)
			{
				m1[i][j]=m1[j][i]=abs(x[i]-x[j]);//暂存6个点之间的距离 
			}
		}
		m1[1][2]=m1[2][1]=q;//把3座桥对应的顶点之间的距离填充到最短路径上 
		m1[3][4]=m1[4][3]=q;//无向图 
		m1[5][6]=m1[6][5]=q;
		floyd();//更新6个点之间的最短路 
		for(k=1;k<=m;k++)
		{
			scanf("%d%d",&h,&l);
			ans=abs(h-l);//待测路径的初始值 
			for(i=1;i<=6;i++)
			for(j=1;j<=6;j++)
			{      
				ans=min(ans,abs(h-x[i])+m1[i][j]+abs(l-x[j])) ; //h与l，x[i]与x[j]的前后关系不能确定，所以需要把h与l的位置互换来确定最小值 
				ans=min(ans,abs(l-x[i])+m1[i][j]+abs(h-x[j])) ;
			}
			ans=(ans*k)%mod;
			sum=(sum+ans)%mod;
		}
		sum%=mod;
		printf("%I64d\n",sum);
	}
	
	return 0;
}